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Because each increment is $\pm 1$, the event at time $k$ defining $T$ is exactly $\{X_{k-1}=X_k=1\}$. Therefore, for $n\ge 2$,and for $n<2$ this event is empty. Thus $T$ is a stopping time for the natural filtration.

On the other hand, $\{U\le 0\}=\{T=2\}=\{X_1=X_2=1\}$. This event has probability $1/4$, whereas the initial sigma-algebra ${\mathcal{F}}_0$ is trivial, even after completion up to null sets. It therefore cannot belong to ${\mathcal{F}}_0$. Hence $U=T-2$ is not a stopping time. Subtracting a deterministic delay from a stopping time can require information from the future.

Use the predictable sigma-algebra generated by $A\times \{0\}$, $A\in {\mathcal{F}}_0$, and $A\times (s,t]$, $A\in {\mathcal{F}}_s$. A previsible process is a measurable function for this sigma-algebra. Set $J_t=1_{\{S<t\le T\}}$, with $J_0=0$. The events $\{S<t\}$ and $\{T<t\}$ are in ${\mathcal{F}}_t$, because they are countable unions of events $\{S\le q\}$ and $\{T\le q\}$ for rational $q<t$. Thus $J$ is an adapted process. Each path is left-continuous, including at the two endpoints, and is zero if $S=T$.

For a direct measurability argument, letEach summand is measurable for the predictable sigma-algebra, since its coefficient is ${\mathcal{F}}_{k{2}^{-n}}$-measurable. At every $t>0$ the sampling time approaches $t$ strictly from below, so left-continuity gives $J_t^{(n)}\to J_t$. ThereforeThe original PDF has a right-closed interval. The open right endpoint in the converted TeX is a transcription error; replacing $(S,T]$ by $(S,T)$ is not valid for arbitrary stopping times.

An elementary predictable process with stopping-time intervals has the form $H={\sum }_{j=0}^{r-1}{h}_j{1}_{({\tau }_j,{\tau }_{j+1}]}$, where the stopping times are ordered and $h_j$ is bounded and ${\mathcal{F}}_{{\tau }_j}$-measurable. Deterministic endpoints give the usual simple predictable process. Define the zero-starting stochastic integral byCommon refinement and telescoping show that this definition is independent of the representation. Use $M_{\infty }$ for an infinite endpoint. The L2-bounded martingale convergence theorem gives convergence of $M_t$ in $L^2$ to $M_{\infty }$ and $M_t=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_t]$.

Put $D_j=M_{{\tau }_{j+1}}-M_{{\tau }_j}$. The optional sampling theorem gives $\mathbb{E}[D_j\mid {\mathcal{F}}_{{\tau }_j}]=0$. For $j<k$, the variable $h_j{D}_j{h}_k$ is ${\mathcal{F}}_{{\tau }_k}$-measurable; hence $\mathbb{E}[h_j{D}_j{h}_k{D}_k]=0$. Applying the same theorem to the assumed martingale $M^2-[M]$ givesConsequently the Itô isometry follows:The unbounded stopping times cause no optional-sampling gap. The $L^2$ terminal representation makes $M$ uniformly integrable and makes the family of stopped $M^2$ uniformly integrable, by conditional Jensen applied to $M_{\infty }^2$. Also $\mathbb{E}[M{]}_{\infty }=\mathbb{E}{M}_{\infty }^2-\mathbb{E}{M}_0^2<\infty$, from monotone convergence theorem and the assumed square-compensator identity. Thus $M^2-[M]$ has the needed uniform integrability, and truncating stopping times and passing in $L^1$ is legitimate. Bounded coefficients are part of the simple-integrand convention; more general coefficients require the displayed square-integrability condition.

Use the shifted filtration ${\mathcal{G}}_s={\mathcal{F}}_{T+s}$. Each $T+s$ is a stopping time, and the optional sampling theorem implies that $M_s^{(T)}=M_{T+s}-M_T$ is a ${\mathcal{G}}_s$-martingale. It is $L^2$-bounded, for example by $\mathbb{E}|M_{T+s}-M_T{|}^2\le 4\mathbb{E}{M}_{\infty }^2$. Its quadratic variation is $[M^{(T)}{]}_s=[M{]}_{T+s}-[M{]}_T$. The shifted process $H_{T+s}$ is continuous and ${\mathcal{G}}_s$-adapted, hence predictable. Its boundedness and the integrability of the bracket make the right-hand stochastic integral well-defined in $L^2$.

For a deterministic partition $0=s_0<\cdots <s_r=t$, use left-endpoint coefficients $H_{T+s_j}$. Both integrals have the same approximating sumOn the original clock this is the integral of the simple process on the stopping-time intervals $(T+s_j,T+s_{j+1}]$; on the shifted clock it is a deterministic-partition integral. As the mesh tends to zero, continuity of $H$ gives pathwise uniform approximation on the random compact interval $[T,T+t]$. The Itô isometry and domination by a constant times the integrable $[M{]}_{\infty }$ give $L^2$ convergence of both sums. Thus stopping-time shift of a stochastic integral yieldsThe left integral includes increments strictly after $T$, so no jump at the starting time is added.

Let $\tau =S_r\wedge T_R$. The stopped logarithm lies between $\log r$ and $\log R$, so the bounded local martingale criterion makes it a true bounded martingale. Also $\tau$ is finite almost surely. Indeed $|B_t{|}^2-2t$ is a martingale, and applying its stopped identity through $t\wedge T_R$ givesmonotone convergence theorem implies $\mathbb{E}{T}_R\le (R^2-1)/2<\infty$, hence $\tau \le T_R<\infty$ almost surely. Boundedness permits dominated convergence theorem in the stopped logarithm as $t\to \infty$. ConsequentlyThe standard results used are Itô's formula, the bounded-local-martingale criterion, the stopped martingale identity at bounded stopping times, and dominated/monotone convergence theorem. At exit the radius is either $r$ or $R$ by continuity. Solving $p\log r+(1-p)\log R=0$ also gives the planar Brownian annulus hitting probability

The sampling rule in the PDF is with replacement. Thus conditional on the current proportion $y=j/n$, the next count is $W\sim \mathrm{Bin}(n,y)$. Put $\Delta =W/n-y$ and $q=y(1-y)$. Direct binomial moments giveThe fourth centered moment of a binomial distribution implies $\mathbb{E}|\Delta {|}^3\le (\mathbb{E}{\Delta }^2\mathbb{E}{\Delta }^4{)}^{1/2}\le C{n}^{-3/2}$. The accelerated generator of the Wright–Fisher binomial sampling chain isFor $f\in C^3([0,1])$, Taylor's formula uniformly yieldsThe initial condition is $Z_0^n=\lfloor xn\rfloor$, so $X_0^n\to x$. The endpoints are absorbing.

We give the path-space argument, rather than only a formal generator calculation. Regard $X^n$ as a càdlàg process in $D([0,1],[0,1])$ with the Skorokhod $J_1$ topology. The chain proportion is a bounded martingale, with predictable square compensator $V_t^n=n^{-1}{\sum }_{k<\lfloor nt\rfloor }{X}_{k/n}^n(1-X_{k/n}^n)$. For bounded stopping times ${\tau }_n\le {\sigma }_n\le {\tau }_n+{\delta }_n$, optional sampling gives $\mathbb{E}|X_{{\sigma }_n}^n-X_{{\tau }_n}^n{|}^2=\mathbb{E}(V_{{\sigma }_n}^n-V_{{\tau }_n}^n)\le ({\delta }_n+n^{-1})/4$. This bound uses the deterministic number of grid jumps in an interval of length at most ${\delta }_n$. Compact containment is automatic. The Aldous tightness criterion therefore proves tightness.

Furthermore, the fourth-moment bound and a union bound giveEvery subsequential limit is continuous. For each smooth $f$, the discrete compensated processis a martingale. Generator convergence and continuity of any limiting path turn the sum into ${\int }_0^t\frac{1}{2}{X}_s(1-X_s)f^{\prime \prime }(X_s)ds$. These compensated processes are uniformly bounded on $[0,1]$, so their martingale identities pass to the limit against bounded continuous functions of past coordinates. A Monotone class theorem argument extends the identities to the natural filtration. Smooth approximation in the $C^2$ norm extends the test class. Thus every limit solves the martingale problem for $Lf(y)=\frac{1}{2}y(1-y)f^{\prime \prime }(y)$ on $[0,1]$.

To identify this problem with the stated SDE, tests equal to $y$ and $y^2$ on $[0,1]$ (using bounded smooth extensions) give a continuous martingale with bracket ${\int }_0^t{X}_s(1-X_s)ds$. On an enlargement carrying independent Brownian motion $W$, setIts bracket is $t$, and the Lévy characterization of Brownian motion makes it Brownian. The zero-coefficient portion of $dX$ has zero bracket, so $d{X}_t=\sqrt{X_t(1-X_t)}d{B}_t$. Conversely Itô's formula shows that any such SDE solution solves the limiting martingale problem. The allowed uniqueness of the SDE, understood at least as uniqueness in law, therefore makes every subsequential limit have the same law. ConsequentlyThe limit is the Wright–Fisher diffusion. Linear interpolation has the same weak limit in the uniform topology because the maximum jump tends to zero. The results applied are Aldous tightness, the vanishing-jump continuous-limit criterion, passage of uniformly bounded martingale identities under weak convergence, Lévy characterization, and uniqueness in law; the estimates above verify their hypotheses.

For an active monster, independence givesBoth probabilities vanish before time $100$ and at odd times. By convolution and Cauchy-Schwarz, $p_{2m}(a,0)\le p_{2m}(0,0)<1/m$: indeed, write the $2m$-step probability as the inner product of two translates of the $m$-step probability mass function. ThereforePositivity follows by prescribing $100$ simultaneous steps taking both walkers straight to the origin; that event has probability $4^{-200}$. An active monster thus leaves a positive escape probability.

For a sleepy monster, the origin capture by a sufficiently sleepy planar random walk can be proved using the printed schedule: during $I_j=\{n_j+1,\dots ,n_{j+1}\}$ it occupies $Y_{j+1}$. For any $M$, let $J_M$ be the first $j\ge M$ with $Y_{j+1}=0$. Markov-chain recurrence of $Y$ makes $J_M$ finite almost surely, and it is independent of $X$. By conditioning on $J_M$ and using the stated block bound,As $M\to \infty$, continuity of probability for decreasing events shows that the event $C_{\infty }$ of infinitely many simultaneous origin visits has probability at least $1/2$. Regard the independent trajectories as continuing even after a possible first capture.

To turn this into probability one, use the eventual coupling of planar symmetric random walks; the block events need not be independent. Two planar simple symmetric random walks begun at sites of the same parity can be coupled to agree eventually. Run them independently until they meet. Their difference walk is irreducible on the even-parity sublattice, and its $m$-step return probability is $p_{2m}(0,0)$, by convolution and symmetry. The preceding divergent-series criterion makes this difference walk a recurrent Markov chain, so it hits zero almost surely. After that stopping time, use identical increments for both copies, preserving their marginal laws by the Strong Markov property.

Fix any finite calendar time $N$ and any two possible histories up to that time. Corresponding princess positions have the same parity, as do the corresponding monster positions at its internal walk time. Apply the eventual probabilistic coupling separately to the two princess futures and the two monster futures, keeping the princess and monster walks independent in each pair. Both the calendar clock and the monster's internal clock tend to infinity. Hence after a finite calendar time the two pairs of trajectories agree, so either both realize $C_{\infty }$ or neither does. The conditional probability of $C_{\infty }$ is therefore the same for every possible finite history.

Consequently $\mathbb{E}(1_{C_{\infty }}\mid {\mathcal{F}}_N)=\mathbb{P}(C_{\infty })$ for every $N$. By the Martingale convergence theorem, these conditional expectations converge to $1_{C_{\infty }}$ as the histories exhaust the trajectories. Thus $\mathbb{P}(C_{\infty })$ is zero or one. Since it is at least $1/2$, it is one. In particular,This argument uses the supplied unconditional block probabilities; it does not silently strengthen them to conditional probabilities given past failures.

Write $\tau ={\tau }_x$ and $T_t=t\wedge \tau$. Path continuity gives $|B_{T_t}|\le x$. Stop the martingale $B_s^2-s$ at the bounded stopping time $T_t$ and apply the optional stopping theorem:By the monotone convergence theorem, $\mathbb{E}\tau \le x^2$, so $\tau <\infty$ almost surely. Path continuity then gives $B_{\tau }\in \{-x,x\}$. The dominated convergence theorem for the bounded variables $B_{T_t}^2$ shows

Next stop the quartic Hermite polynomial martingale from part (a), again only at $T_t$. Its expectation is zero, soMonotone convergence proves $\mathbb{E}{\tau }^2\le 2x^4$, establishing the needed second-moment integrability before the final passage to the limit. Since $T_t{B}_{T_t}^2\le x^2\tau$ and $B_{T_t}^4\le x^4$, dominated convergence givesConsequently the Brownian symmetric interval-exit moments are

To obtain the Laplace transform of symmetric Brownian interval-exit time, put $a=\sqrt{2\lambda }$. The Exponential martingale for Brownian motion shows thatis a martingale with $H_0=1$. Bounded-time stopping gives $\mathbb{E}{H}_{T_t}=1$. Its stopped values are bounded by $\cosh (ax)$, so dominated convergence applies as $t\to \infty$. Since $\cosh (a{B}_{\tau })=\cosh (ax)$, it yieldsEvery use of stopping at $\tau$ has thus been justified through bounded stopping and an explicit integrability or domination argument.

Fix $t$ and put $\sigma =t\wedge \tau$. The stopping time property implies that $\sigma$ is an ${\mathcal{F}}_t$-measurable random variable with values in $[0,t]$: for $s<t$,and for $s\ge t$ the event is the whole space. By part (c), the restriction of $X$ to $[0,t]\times \Omega$ is $\mathcal{B}([0,t])\otimes {\mathcal{F}}_t$-measurable.

The evaluation map $\omega \mapsto (\sigma (\omega ),\omega )$ is measurable from $(\Omega ,{\mathcal{F}}_t)$ to this product space: the inverse image of a measurable rectangle $A\times C$ is $\{\sigma \in A\}\cap C$. Composing it with the jointly measurable stochastic process gives an ${\mathcal{F}}_t$-measurable random variableSince this holds for every fixed $t$, the stopped process is adapted. If path regularity holds only almost surely, first apply the proof to its pathwise regular representative; with a completed filtration, the original stopped variable differs only on a null event and is also ${\mathcal{F}}_t$-measurable. The adaptedness of a stopped right-continuous process requires neither boundedness of $\tau$ nor a martingale assumption; $\tau =\infty$ is harmless because $t\wedge \tau \le t$.

The Skorokhod embedding of a centered random walk states that a random walk with independent identically distributed centered steps of finite variance ${\sigma }^2$ can be realized on an appropriate probability space aswhere $B$ is a standard Brownian motion and the $T_n$ are finite stopping times. More precisely, the stopped positions have the same joint law as the given random walk, and the pairsmay be chosen independent and identically distributed. Their spatial component has the step law, and $\mathbb{E}(T_n-T_{n-1})={\sigma }^2$. Repeating the one-step Skorokhod embedding theorem with the Strong Markov property gives this formulation. In the present normalization, the mean time increment is one, and the strong law of large numbers gives $T_n/n\to 1$ almost surely.

The Donsker invariance principle states that the linearly interpolated diffusively rescaled random walkconverges weakly as a random element of $C[0,1]$, equipped with the uniform norm, to standard Brownian motion restricted to $[0,1]$. At $t=1$ the fractional term is zero. The only step assumptions needed here are zero mean, unit variance, and independent identical distributions; a higher moment or bounded support is not required. This is a functional central limit theorem, concerning the entire interpolated path rather than only its endpoint.

Continuity and adaptedness make the first boundary hit a stopping time. A continuous path starting in the open domain cannot leave it before meeting its boundary. Thus $X_{t\wedge T}\in \mathcal{D}\cup \partial \mathcal{D}$, with the usual interpretation when $T=\infty$. If $|u|\le K$ on this set, thenThe stopped process is a bounded local martingale; the bounded local martingale criterion makes it a martingale and, in fact, a uniformly integrable martingale. The Martingale convergence theorem gives almost sure and $L^1$ convergence as $t\to \infty$.

Its limit can also be identified pathwise. On $\{T<\infty \}$ the stopped process is eventually constant at $e^{-\lambda T}u(X_T)$. On $\{T=\infty \}$ its absolute value is at most $K{e}^{-\lambda t}$ and hence tends to zero. Thus the limit is $e^{-\lambda T}u(X_T)1_{\{T<\infty \}}$.

Backward induction makes the Snell envelope integrable and adapted: $|U_t|\le |Y_t|+\mathbb{E}(|U_{t+1}|\mid {\mathcal{F}}_t)$. Its definition gives $U_t\ge Y_t$ and $U_t\ge \mathbb{E}(U_{t+1}\mid {\mathcal{F}}_t)$, so it is a supermartingale dominating the reward.

For a stopping time $\tau$ taking values in $\{0,\dots ,T\}$, expand its stopped value asThe indicators are ${\mathcal{F}}_t$-measurable. Taking conditional expectations in each summand makes its expectation nonpositive by the supermartingale property. Since ${\mathcal{F}}_0$ is trivial, $U_0$ is deterministic andThis proves the finite-horizon optional sampling theorem directly in the instance needed here, without assuming nonnegative rewards.

The first-contact time ${\tau }_{\ast }\le T$ is a stopping time: the event $\{{\tau }_{\ast }\le t\}$ is the finite union of events $\{U_s=Y_s\}$ for $s\le t$. It is finite because $U_T=Y_T$. On $\{{\tau }_{\ast }>t\}$ we have $U_t>Y_t$, so the maximum defining the Snell envelope selects continuation andThe stopped process $U_{t\wedge {\tau }_{\ast }}$ is therefore a martingale: before stopping its conditional increment is zero and after stopping its increment vanishes. Applying the stopped-sum argument with zero conditional increments givesThe upper bound from part (a) now proves that first contact is an optimal stopping time.

Use the natural filtration ${\mathcal{F}}_n=\sigma (X_1,\dots ,X_n)$. The first exit time is a stopping time. If $r\le x$, it is zero, so its expectation is already finite. Now assume $0<x<r$.

The positive-increment hypothesis gives a number $h>0$ with $\mathbb{P}(X_1>h)>0$. Choose an integer $m$ so large that $r/m<h$, and put $p=\mathbb{P}(X_1>r/m)>0$. A block of $m$ increments all exceeding $r/m$ has probability $p^m$. From any point still in $(0,r)$, that block forces an upper exit before the block ends.

By independence, conditional on ${\mathcal{F}}_{km}$ and survival to time $km$, the next block has this same probability. Thus the geometric tail bound from a uniform escape probability givesUsing the tail-sum formula for the expected value of a nonnegative integer-valued random variable,This is the random-walk exit bound from a positive-increment block. In particular, the exit occurs with probability one. The mean-zero assumption is not needed for this first bound.

Let $T\le N$ for a deterministic integer $N$. The stopped martingale has the finite-sum representationEach summand is integrable, and $\{T\ge k\}=\{T>k-1\}\in {\mathcal{F}}_{k-1}$ by the stopping time property. Therefore the conditional expectation identity for a martingale givesTaking expectations in the finite sum proves the bounded optional stopping theorem:No limiting argument or uniform-integrability assumption is needed for a bounded stopping time.

The bounded optional stopping theorem gives $\mathbb{E}{X}_{T\wedge n}=\mathbb{E}{X}_0$. Decompose the difference from the terminal value:ConsequentlyThe first term tends to zero by hypothesis. The second tends to zero by the dominated convergence theorem, because $X_T$ is integrable and $T$ is finite with probability one. Thus the stopped martingale converges to $X_T$ with convergence in L1, which permits passage of expectations to the limit:The explicit tail condition supplies exactly the missing control for an unbounded stopping time.